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Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 m to the right of antenna \(A\). Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna \(B\). The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q\)? (b) What is the longest wavelength for which there will be constructive interference at point \(Q\)?

Short Answer

Expert verified
The longest wavelength for destructive interference is 240 m, and for constructive interference is 120 m.

Step by step solution

01

Understanding the Problem

We need to determine the conditions for both destructive and constructive interference at the point Q on the line that extends through the two radio antennas, A and B. Given the distances, antenna B is 120 m from antenna A, and point Q is 40 m from antenna B.
02

Calculate the Path Difference

The total distance from antenna A to point Q is 120 m + 40 m = 160 m. The distance from antenna B to point Q is 40 m. The path difference, \( \Delta d \), is the difference between these two distances: \( \Delta d = 160 \text{ m} - 40 \text{ m} = 120 \text{ m} \).
03

Determine Condition for Destructive Interference

For destructive interference, the path difference should be an odd multiple of half the wavelength, i.e., \( \Delta d = \left( n + \frac{1}{2} \right) \lambda \), where \( n \) is an integer (0, 1, 2, ...). The longest wavelength occurs when \( n = 0 \). Setting \( n = 0 \) gives: \( 120 = \frac{1}{2} \lambda \), so \( \lambda = 240 \) m.
04

Determine Condition for Constructive Interference

For constructive interference, the path difference should be a multiple of the wavelength, i.e., \( \Delta d = m \lambda \), where \( m \) is an integer. The longest wavelength occurs when \( m = 1 \). Using \( \Delta d = 120 \) m gives: \( 120 = 1 \cdot \lambda \), so \( \lambda = 120 \) m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive Interference
When waves meet, their effects can either cancel each other out or amplify. Destructive interference happens when the peak of one wave meets the trough of another, leading them to cancel.
The result is often a reduction in wave amplitude or, in the case of identical waves, complete cancellation.
For destructive interference to occur, the path difference (\(\Delta d\)) between two waves must be an odd multiple of half the wavelength (\(\lambda\)).
  • The formula for the condition is \[\Delta d = \left(n + \frac{1}{2}\right) \lambda\], where \(n\) is an integer (0, 1, 2, ...).
  • The longest wavelength for destructive interference is when \(n = 0\), simplifying the formula to \(\Delta d = \frac{\lambda}{2}\).
In our problem, \(\Delta d\) is 120 m, leading to \(\lambda = 240\) m when \(n = 0\).This means point \(Q\) experiences destructive interference at this wavelength.
Constructive Interference
Constructive interference occurs when waves meet in such a way that their crests and troughs align.
This alignment leads the wave amplitudes to add together, resulting in a stronger and more pronounced wave.
The condition for constructive interference is met when the path difference (\(\Delta d\)) is a whole number multiple of the wavelength (\(\lambda\)).
  • Expressed mathematically as \[\Delta d = m \lambda\], where \(m\) is an integer (1, 2, 3, ...).
  • The longest wavelength occurs when \(m = 1\), as it represents the fundamental mode of interference.
In our scenario, with \(\Delta d = 120\) m, the longest wavelength for constructive interference is \(\lambda = 120\) m, making point \(Q\) a zone of amplification at this wavelength.
Path Difference
Path difference is critical in understanding wave interference. It refers to the difference in distance traveled by two waves from their sources to a common point.
In terms of interference, this difference determines whether waves will constructively or destructively interfere at that point.
The concept can be explained as follows:
  • For constructive interference, the path difference should be a multiple of the wavelength: \(m \lambda\).
  • For destructive interference, it should be an odd multiple of half the wavelength: \(\left(n+\frac{1}{2}\right) \lambda\).
In the given exercise, the path difference from antennas \(A\) and \(B\) to point \(Q\) is calculated as 120 m (160 m - 40 m).
This path difference guides us in determining the conditions for both constructive and destructive interference.
Wavelength Calculation
Calculating the wavelength is key to understanding interference patterns. Wavelength (\(\lambda\)) can be found using the interference conditions based on the path difference.
The method relies on understanding how the path difference relates to \(\lambda\):
  • For constructive interference, use the formula \(\Delta d = m \lambda\)
  • For destructive interference, use \(\Delta d = \left(n + \frac{1}{2}\right) \lambda\)
By plugging the path difference and the required condition's integer (\(m\) or \(n\)) into these formulas, we can solve for \(\lambda\).
In our problem, the given path difference of 120 m helps determine wavelengths for both types of interference:
  • For constructive interference: \(\lambda = 120\) m (\(m = 1\)
  • For destructive interference: \(\lambda = 240\) m (\(n = 0\)
This comprehensive understanding assists in visualizing how waves interact at different points.

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Most popular questions from this chapter

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.

What is the thinnest film of a coating with \(n\) = 1.42 on glass (\(n\) = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_0\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0\(^\circ\), what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase difference is 60.0\(^\circ\)?

Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)

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